IMCA mechanical calculator

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An italian calculator designed around 1958.

Inside Operation Add and Subtract Multiply and Divide Square root



Inside
This calculator can perform the four basic operations :
  • Operands with 10 digits.
  • Accumulator with 13 digits.
  • Quotient with 8 digits.
  • Second factor, when multiplying, with 8 digits.
  • Copy back from accumulator to Operand.
  • Bell ring when overflow.

Cleaning the machine.


Operation
  • To input an operand, use the cursors B, the number appears in A. The lever L sets the operand to zero.
  • The lever K adds or substracts (according to the rotation) the operand in A to the accumulator in C.
  • Each rotation is counted in D, the counter. Its digits are white for additions and red for substractions.
  • The lever E set the counter back to zero and, if F is in the low position, it also resets the accumulator C.
  • The accumulator C can be switched left or right compared with A to sum directly on tens, hundreds, etc. The lever I/J switches the carrier one position left or right. For a free and fast move, use the lever H.
  • To copy C back to A, lower G (and F) and use the lever E to clear the accumulator as you copy it to A. Then, don't forget to raise G with a small pin on the right side to free every other movement.
  • The cursors M are free and you can use them to set the comma.


Add and Substract


Let's calculate 1524 + 97


Note Actions Operand A/B Counter D Accumulator C
Clear all F- E L
0000000000
00000000
0000000000000
Set first number and add it to the accumulator B=1524 K+
0000001524
00000001
0000000001524
Set second number and add it! L B=97 K+
0000000097
00000002
0000000001621


Then 1524 + 97 = 1621




Let's calculate 20.14-19.66


Note Actions Operand A/B Counter D Accumulator C
Clear all, comma to second digit F- E L M2
0000000000
00000000
0000000000000
Set first number and add it to the accumulator B=2014 K+
0000002014
00000001
0000000002014
Set second number and substract it! L B=1966 K-
0000001966
00000000
0000000000048


Then 20.14 - 1966 = 0.48




Multiply and Divide


How many hours in a year? 365 * 24


Note Actions Operand A/B Counter D Accumulator C
Clear all F- E L
0000000000
00000000
0000000000000
Set first number and add it 4 times to the accumulator B=365 K+ K+ K+ K+
0000000365
00000004
0000000001460
Shift right to the tens, and add the number twice to the accumulator I K+ K+
0000000365
00000024
0000000008760


Then, there are 8760 hours in one year.




Sums and differences of products (314 * 12) + (15 * 24) - (47 * 31)


It's easy, each product is computed with the previous method, but, between them, you just clear the counter and not the accumulator! (Position F+ when using the lever E). The effect is that they are simply added!
If you want to substract, just rotate the other way with K-.


Approximate PI with the quotient 22 / 7 with three decimals.


The result will appear in Counter D, it's the number of turns that gives the quotient!

Note Actions Operand A/B Counter D Accumulator C
Clear all, comma to third digit for C and D F- E L M3
0000000000
00000000
0000000000000
Set first number, shift carrier 3 positions to the right, add it to the accumulator B=22 I I I K+
0000000022
00001000
0000000022000
Reset counter and set the divisor F+ E B=7
0000000007
00000000
0000000022000
Substract 7 from 22 as many times as possible (*), shift left one digit K- K- K- J
0000000007
00003000
0000000001000
Substract 7 from 10 as many times as possible, shift left one digit K- J
0000000007
00003100
0000000000300
Substract 7 from 30 as many times as possible, shift left one digit K- K- K- K- J
0000000007
00003140
0000000000020
Substract 7 from 20 as many times as possible, shift left one digit K- K- J
0000000007
00003142
0000000000006


Then PI is close to 3.142


(*) Note : if you substract too many times, then you hear a bell ring indicating an overflow. Then, just turn it back with K+, you'll hear another bell ring!

Square root


Let's calculate the square root of 215.4


You have to cut this number in blocks of two digits starting from the comma (because 100 = 10²):
02 15.40
We'll start working on the first block, the 2. On my IMCA, to get a good precision, place the carrier to have the arrow pointing at the 7th digit in the counter D. The, we'll get 6 or 7 digits for the square root acording to the number of digits in the first block.

This method uses the fact that the sum of the odd numbers is a square (1+3+5=9=3² for example). More details and theory here.

Note Actions Operand A/B Counter D Accumulator C
Clear all, columns B7-B4 we program the value and send it to the accumulatorr F- E L M3 B=2154 K+
0002154000
01000000
2154000000000
Clear the counter, column of the first block (B7) start with 1 F+ E L B=1
0001000000
00000000
2154000000000
Substract the odd numbers in serie until a bell ring, then turn back once, decrement B and move carrier left! B=1 K- B=3 K- ♪ K+ ♪ B=2 J
0002000000
01000000
1154000000000
Columns B7-B6 program the next odd number, 21, Substract the odd numbers in serie until a bell ring, then turn back once, decrement B and move carrier left! B=21 K- B=23 K- B=25 K- B=27 K- B=29 K- ♪ K+ ♪ B=28 J
0002800000
01400000
0194000000000
Columns B7-B5 program the next odd number, 281, Substract the odd numbers in serie until a bell ring, then turn back once, decrement B and move carrier left! B=281 K- B=283 K- B=285 K- B=287 K- B=289 K- B=291 K- B=293 K- ♪ K+ ♪ B=292 J
0002920000
01460000
0022400000000
Then we would go on, columns B7-B4, with the odd value B=2921...

Then, the square root of 215.4 is close to 14.6




Depuis le 15 décembre 2007